*Sửa đề: "35 gam CaO"
PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\)
a) Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{120}{100}=1,2\left(mol\right)=n_{CaO\left(lý.thuyết\right)}\\n_{CaO\left(thực\right)}=\dfrac{35}{56}=0,625\left(mol\right)\end{matrix}\right.\) \(\Rightarrow H\%=\dfrac{0,625}{1,2}\cdot100\%\approx52,08\%\)
b) Với hiệu suất 75% \(\Rightarrow n_{CaO\left(thực\right)}=1,2\cdot75\%=0,9\left(mol\right)\) \(\Rightarrow m_{CaO}=0,9\cdot56=50,4\left(g\right)\)